Rectifiers — Half Wave and Full Wave
Half-Wave Rectifier Average Voltage
V_{dc} = \frac{V_m}{\pi}
| Symbol | Description | Unit |
|---|---|---|
| V_{dc} | Average (DC) output voltage | V |
| V_m | Peak input voltage | V |
Worked example
Half-wave rectifier connected to 230 V rms supply. Find Vdc.
Given: Vrms=230 V
- Vm = Vrms × √2 = 230 × 1.414 = 325.3 V
- Vdc = Vm/π = 325.3/3.1416 = 103.6 V
Answer: Vdc = 103.6 V
Full-Wave Rectifier Average Voltage
V_{dc} = \frac{2 V_m}{\pi}
| Symbol | Description | Unit |
|---|---|---|
| V_{dc} | Average output voltage | V |
| V_m | Peak input voltage | V |
Worked example
Full-wave bridge rectifier, 230 V rms input. Find Vdc and Vrms output.
Given: Vrms_in=230 V
- Vm = 230√2 = 325.3 V
- Vdc = 2×325.3/π = 650.6/3.1416 = 207.1 V
- Vrms_out = Vm/√2 = 325.3/1.414 = 230 V (full wave)
Answer: Vdc = 207.1 V, Vrms_out = 230 V
Ripple Factor
RF = \frac{V_{rms,ac}}{V_{dc}} = \sqrt{\left(\frac{V_{rms}}{V_{dc}}\right)^2 - 1}
| Symbol | Description | Unit |
|---|---|---|
| RF | Ripple factor | |
| V_{rms,ac} | RMS of AC ripple component | V |
| V_{dc} | DC component | V |
Worked example
Full-wave rectifier: Vdc=207.1 V, Vrms=230 V. Find ripple factor.
Given: Vdc=207.1 V, Vrms=230 V
- RF = √((Vrms/Vdc)² − 1) = √((230/207.1)² − 1)
- = √((1.1106)² − 1) = √(1.2334 − 1) = √0.2334 = 0.483
Answer: RF = 0.483 (full-wave, no filter)
Rectifier Efficiency
\eta = \frac{V_{dc}^2 / R_L}{V_{rms}^2 / R_L} = \left(\frac{V_{dc}}{V_{rms}}\right)^2 \times 100\%
| Symbol | Description | Unit |
|---|---|---|
| \eta | Rectification efficiency | % |
| V_{dc} | Average output voltage | V |
| V_{rms} | RMS output voltage | V |
Worked example
Full-wave rectifier: Vdc=207.1 V, Vrms=230 V. Find efficiency.
Given: Vdc=207.1, Vrms=230
- η = (207.1/230)² × 100
- = (0.9004)² × 100 = 0.8107 × 100 = 81.07%
Answer: η = 81.07%
Thyristor (SCR) Controlled Rectifiers
Single-Phase Half-Controlled Average Output Voltage
V_{dc} = \frac{V_m}{\pi}(1 + \cos\alpha)
| Symbol | Description | Unit |
|---|---|---|
| V_{dc} | Average output voltage | V |
| V_m | Peak supply voltage | V |
| \alpha | Firing angle | ° |
Worked example
Single-phase half-controlled rectifier: Vm=325 V, α=60°. Find Vdc.
Given: Vm=325 V, α=60°
- Vdc = (325/π)(1+cos60°)
- = (325/3.1416)(1+0.5)
- = 103.45 × 1.5 = 155.2 V
Answer: Vdc = 155.2 V
Three-Phase Fully-Controlled Rectifier Output
V_{dc} = \frac{3\sqrt{3}}{\pi} V_m \cos\alpha = 2.34 V_{LL,rms} \cos\alpha
| Symbol | Description | Unit |
|---|---|---|
| V_{dc} | Average DC output | V |
| V_m | Peak phase voltage | V |
| V_{LL,rms} | Line-to-line RMS voltage | V |
| \alpha | Firing angle | ° |
Worked example
3-phase fully controlled rectifier: VLL=415 V, α=30°. Find Vdc.
Given: VLL=415 V, α=30°
- Vdc = 2.34 × 415 × cos(30°)
- = 2.34 × 415 × 0.866
- = 2.34 × 359.4 = 840.9 V
Answer: Vdc = 840.9 V
SCR Turn-Off Time Requirement
t_{off} \leq t_c = \frac{\pi - \alpha - \mu}{\omega}
| Symbol | Description | Unit |
|---|---|---|
| t_c | Circuit turn-off time available | s |
| \mu | Overlap angle | rad |
| \alpha | Firing angle | rad |
Worked example
Single-phase converter: α=60°, overlap µ=10°, f=50 Hz. Find circuit turn-off time.
Given: α=60°=π/3, µ=10°=π/18, ω=2π×50=314.16 rad/s
- tc = (π − π/3 − π/18) / 314.16
- = (18π/18 − 6π/18 − π/18) / 314.16
- = (11π/18) / 314.16
- = 1.9199 / 314.16 = 6.11 ms
Answer: tc = 6.11 ms
DC–DC Converters (Choppers)
Buck Converter Output Voltage
V_o = D \cdot V_s
| Symbol | Description | Unit |
|---|---|---|
| V_o | Output voltage | V |
| D | Duty cycle (0 to 1) | |
| V_s | Input supply voltage | V |
Worked example
Buck converter: Vs=24 V, required Vo=5 V. Find duty cycle.
Given: Vs=24 V, Vo=5 V
- D = Vo/Vs = 5/24 = 0.2083
- Duty cycle ≈ 20.83%
Answer: D = 0.2083
Boost Converter Output Voltage
V_o = \frac{V_s}{1 - D}
| Symbol | Description | Unit |
|---|---|---|
| V_o | Output voltage (always > Vs) | V |
| D | Duty cycle | |
| V_s | Input voltage | V |
Worked example
Boost converter: Vs=12 V, D=0.4. Find output voltage.
Given: Vs=12 V, D=0.4
- Vo = 12/(1−0.4) = 12/0.6 = 20 V
Answer: Vo = 20 V
Buck-Boost Converter Output Voltage
V_o = -\frac{D}{1-D} V_s
| Symbol | Description | Unit |
|---|---|---|
| V_o | Output voltage (inverted polarity) | V |
| D | Duty cycle |
Worked example
Buck-boost converter: Vs=15 V, D=0.5. Find output.
Given: Vs=15 V, D=0.5
- Vo = −(0.5/(1−0.5))×15 = −(0.5/0.5)×15 = −1×15 = −15 V
Answer: Vo = −15 V (inverted, same magnitude)
Output Ripple Voltage (Buck Converter)
\Delta V_o = \frac{V_o(1-D)}{8LCf^2}
| Symbol | Description | Unit |
|---|---|---|
| \Delta V_o | Peak-to-peak output ripple | V |
| L | Inductor value | H |
| C | Output capacitor | F |
| f | Switching frequency | Hz |
Worked example
Buck: Vo=5 V, D=0.208, L=100 µH, C=100 µF, f=50 kHz. Find output ripple.
Given: Vo=5, D=0.208, L=100×10^−6, C=100×10^−6, f=50×10^3
- ΔVo = 5×(1−0.208)/(8×100×10^−6×100×10^−6×(50×10^3)²)
- = 5×0.792 / (8×10^−10×2.5×10^9)
- = 3.96 / 2 = 1.98 V
Answer: ΔVo = 1.98 V
Inverters — Single-Phase
Single-Phase Full-Bridge Inverter Fundamental Output
V_{o1} = \frac{4 V_{dc}}{\pi} \sin\left(\frac{m\pi}{2}\right)
| Symbol | Description | Unit |
|---|---|---|
| V_{o1} | Fundamental component of output voltage | V |
| V_{dc} | DC bus voltage | V |
| m | Modulation index (0 to 1 for linear range) |
Worked example
Single-phase full-bridge inverter: Vdc=300 V. Find peak fundamental output (square wave, m=1).
Given: Vdc=300 V, m=1
- Vo1_peak = 4×300/π × sin(π/2) = 1200/π × 1
- = 1200/3.1416 = 381.97 V
- Vrms_fundamental = 381.97/√2 = 270 V
Answer: Peak = 381.97 V; Vrms = 270 V
PWM Modulation Index
m_a = \frac{\hat{V}_{control}}{\hat{V}_{triangle}}
| Symbol | Description | Unit |
|---|---|---|
| m_a | Amplitude modulation index | |
| \hat{V}_{control} | Peak of reference (sinusoidal) signal | V |
| \hat{V}_{triangle} | Peak of carrier (triangular) signal | V |
Worked example
SPWM: control signal peak = 8 V, carrier peak = 10 V. Find ma and peak output fundamental.
Given: V_ctrl=8 V, V_tri=10 V, Vdc=400 V
- ma = 8/10 = 0.8
- Vo1_peak = ma × Vdc = 0.8 × 400 = 320 V (for single-phase full bridge)
- Vrms = 320/√2 = 226.3 V
Answer: ma = 0.8, Vo1_rms = 226.3 V
THD of Inverter Output
THD = \frac{\sqrt{V_{rms}^2 - V_1^2}}{V_1}
| Symbol | Description | Unit |
|---|---|---|
| THD | Total harmonic distortion | |
| V_{rms} | Total RMS output voltage | V |
| V_1 | Fundamental RMS voltage | V |
Worked example
Inverter output: Vrms=230 V, fundamental V1=220 V. Find THD.
Given: Vrms=230 V, V1=220 V
- THD = √(230² − 220²)/220
- = √(52900 − 48400)/220
- = √4500/220 = 67.08/220 = 0.305 = 30.5%
Answer: THD = 30.5%
AC Voltage Controllers and Cycloconverters
Single-Phase AC Voltage Controller RMS Output
V_{rms} = V_s \sqrt{\frac{1}{\pi}\left[\pi - \alpha + \frac{\sin 2\alpha}{2}\right]}
| Symbol | Description | Unit |
|---|---|---|
| V_{rms} | RMS output voltage | V |
| V_s | RMS supply voltage | V |
| \alpha | Firing angle | rad |
Worked example
AC voltage controller: Vs=230 V, α=60°=π/3. Find Vrms output.
Given: Vs=230 V, α=π/3 rad
- sin(2α) = sin(120°) = 0.866
- Term = π − π/3 + 0.866/2 = 3.1416−1.0472+0.433 = 2.527
- Vrms = 230×√(2.527/π) = 230×√0.8044 = 230×0.8969 = 206.3 V
Answer: Vrms = 206.3 V
MOSFET and IGBT Switching Losses
Switching Loss per Cycle
E_{sw} = \frac{1}{2} V_{ds} I_D (t_r + t_f)
| Symbol | Description | Unit |
|---|---|---|
| E_{sw} | Energy lost per switching event | J |
| V_{ds} | Voltage across device during switching | V |
| I_D | Current through device | A |
| t_r, t_f | Rise and fall times | s |
Worked example
IGBT: Vds=400 V, ID=10 A, tr=100 ns, tf=150 ns, f=20 kHz. Find average switching power.
Given: Vds=400, ID=10, tr=100×10^−9, tf=150×10^−9, f=20000
- E_sw = 0.5×400×10×(100+150)×10^−9
- = 0.5×400×10×250×10^−9
- = 0.5×10^6×10^−6 = 0.5×250×10^−6 = 5×10^−4 J = 0.5 mJ
- P_sw = E_sw × f = 0.5×10^−3 × 20000 = 10 W
Answer: Average switching power = 10 W
Conduction Loss
P_{cond} = I_{rms}^2 \cdot R_{ds(on)} \quad (\text{MOSFET})
| Symbol | Description | Unit |
|---|---|---|
| P_{cond} | Conduction power loss | W |
| I_{rms} | RMS current through device | A |
| R_{ds(on)} | On-state drain-source resistance | Ω |
Worked example
MOSFET: Irms=5 A, Rds(on)=0.2 Ω. Find conduction loss.
Given: Irms=5 A, Rds_on=0.2 Ω
- P_cond = 5² × 0.2 = 25 × 0.2 = 5 W
Answer: P_cond = 5 W
Drives — Speed Control of DC and AC Motors
DC Drive Speed from Converter Output
N = \frac{V_{dc} - I_a R_a}{K \phi}
| Symbol | Description | Unit |
|---|---|---|
| N | Motor speed | rpm |
| V_{dc} | Converter average output voltage | V |
| K\phi | Motor voltage constant × flux | V/rpm |
Worked example
Phase-controlled rectifier supplies DC motor: Vdc=180 V, Ia=20 A, Ra=0.5 Ω, Kφ=0.05 V/rpm. Find speed.
Given: Vdc=180, Ia=20, Ra=0.5, Kφ=0.05
- Eb = Vdc − Ia×Ra = 180 − 20×0.5 = 180−10 = 170 V
- N = Eb/(Kφ) = 170/0.05 = 3400 rpm
Answer: N = 3400 rpm
V/f Control (Induction Motor)
\frac{V}{f} = \text{constant} \Rightarrow V_{new} = V_{rated} \cdot \frac{f_{new}}{f_{rated}}
| Symbol | Description | Unit |
|---|---|---|
| V_{new} | New supply voltage for new frequency | V |
| f_{new} | New supply frequency | Hz |
| f_{rated} | Rated frequency | Hz |
Worked example
Induction motor: rated 415 V, 50 Hz. Run at 30 Hz with constant V/f. Find required voltage.
Given: Vrated=415 V, frated=50 Hz, fnew=30 Hz
- V_new = 415 × (30/50) = 415 × 0.6 = 249 V
Answer: V_new = 249 V
Quick reference
| Formula | Expression |
|---|---|
| HW Rectifier Vdc | V_{dc} = V_m/\pi |
| FW Rectifier Vdc | V_{dc} = 2V_m/\pi |
| Ripple Factor | RF = \sqrt{(V_{rms}/V_{dc})^2-1} |
| Half-Controlled SCR Vdc | V_{dc} = (V_m/\pi)(1+\cos\alpha) |
| 3-Phase Fully Controlled Vdc | V_{dc} = 2.34 V_{LL}\cos\alpha |
| Buck Vo | V_o = D \cdot V_s |
| Boost Vo | V_o = V_s/(1-D) |
| Buck-Boost Vo | V_o = -D V_s/(1-D) |
| Buck Ripple | \Delta V_o = V_o(1-D)/(8LCf^2) |
| Inverter Fundamental | V_{o1} = 4V_{dc}/\pi |
| SPWM ma | m_a = V_{ctrl}/V_{tri} |
| THD | THD = \sqrt{V_{rms}^2-V_1^2}/V_1 |
| Switching Loss | E_{sw} = 0.5 V I (t_r+t_f) |
| Conduction Loss | P_{cond} = I_{rms}^2 R_{ds(on)} |
| V/f Control | V_{new} = V_{rated} \cdot f_{new}/f_{rated} |
Exam tips
- GATE power electronics questions always specify firing angle α in degrees but formulas use radians — convert before calculating; a 30° error costs 0.5–1 marks each time.
- Buck, boost, and buck-boost converters are compared in a single GATE question every few years — memorise the table: Vo/Vs = D, 1/(1-D), and -D/(1-D) respectively.
- Rectifier efficiency (81% FW, 40.5% HW) and ripple factor values (0.48 FW, 1.21 HW) are sometimes asked as MCQs — memorise these standard values.
- SCR commutation overlap questions require knowing that increasing firing angle α reduces the available turn-off time; α near 90° is the critical design limit.
- IGBT switching loss and conduction loss questions test whether students add both losses for total device dissipation — examiners give partial credit if only one component is computed.
- V/f control problems are straightforward ratio calculations but examiners add a trap: above base speed, voltage is clamped at rated and only frequency increases — recognise this regime.